Problem: A big football championship will occur soon! n teams will compete in it, and each pair of teams will play exactly one game against each other.

There are two possible outcomes of a game:

  * the game may result in a tie, then both teams get 1 point; 
  * one team might win in a game, then the winning team gets 3 points and the losing team gets 0 points. 



The score of a team is the number of points it gained during all games that it played.

You are interested in a hypothetical situation when all teams get the same score at the end of the championship. A simple example of that situation is when all games result in ties, but you want to minimize the number of ties as well.

Your task is to describe a situation (choose the result of each game) so that all teams get the same score, and the number of ties is the minimum possible.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.

Then the test cases follow. Each test case is described by one line containing one integer n (2 ≤ n ≤ 100) — the number of teams.

Output

For each test case, print (n(n - 1))/(2) integers describing the results of the games in the following order: the first integer should correspond to the match between team 1 and team 2, the second — between team 1 and team 3, then 1 and 4, ..., 1 and n, 2 and 3, 2 and 4, ..., 2 and n, and so on, until the game between the team n - 1 and the team n.

The integer corresponding to the game between the team x and the team y should be 1 if x wins, -1 if y wins, or 0 if the game results in a tie.

All teams should get the same score, and the number of ties should be the minimum possible. If there are multiple optimal answers, print any of them. It can be shown that there always exists a way to make all teams have the same score.

Example

Input


2
2
3


Output


0 
1 -1 1 

Note

In the first test case of the example, both teams get 1 point since the game between them is a tie.

In the second test case of the example, team 1 defeats team 2 (team 1 gets 3 points), team 1 loses to team 3 (team 3 gets 3 points), and team 2 wins against team 3 (team 2 gets 3 points).

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define reg register
typedef long long ll;
using namespace std;
inline int qr(){
	int x=0,f=0;char ch=0;
	while(!isdigit(ch)){f|=ch=='-';ch=getchar();}
	while(isdigit(ch)){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
	return f?-x:x;
}
const int maxn=110;
int n;
int o[maxn]; 
int c[maxn],d[maxn];
int main(){
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	int t=qr();
	while(t--){
		n=qr();
		int p=(n-1)/2;
		memset(o,0,sizeof(o)); 
		memset(d,0,sizeof(d));
		memset(c,0,sizeof(c));
		if(n&1){
			for(reg int i=1;i<n;i++){
				for(reg int j=i+1;j<=n;j++){
					if(o[i]<p){
						o[i]++;printf("%d ",1);
					}else if(o[j]<p){
						o[j]++;printf("%d ",-1);
					}
				}
			}
				
		}
		else {//偶数才会出现(n-1)/2出现打平 
			for(reg int i=1;i<n;i++){
				for(reg int j=i+1;j<=n;j++){
					if(!d[i]&&!d[j]){
						printf("%d ",0);
						d[i]=d[j]=1;
					}else{
						if(o[i]<p&&c[j]<p){
						c[j]++;
						o[i]++;printf("%d ",1);
						}else if(o[j]<p&&c[i]<p){
						c[i]++;
						o[j]++;printf("%d ",-1);
						}
					}
				}
			}
		}
		printf("\n");
	} 
	return 0;
}
/*
平局尽量少并且使得队伍得分最后相等 
n次比赛中即可产生一批相等,每个数都会有n-1次比赛,如果n-1是偶数,那么每个人都可以做到一半失败一般胜利
如果n-1是奇数,那么只要n-2次输赢,然后最多有n次平局即可,可以用一个桶记录每个数的胜负,当这个数的胜利同达到n-1/2的时候就收手
另外,当这个数不着急胜利的时候(它还有机会胜利使得胜利次数为n-1),那么右边这个数如果着急胜利(他已经失败了n-1次),那么就一定要让后者胜利 
平手的时候是唯一的！因为它也加分！也就是说当遇到一次x,y不同的时候一定要直接打平,这样最后得分才相同； 
*/